Solving a First ODE Using an Integrating Factor

In summary: Try again...In summary, to solve this first-order DE, you can use the method of separation by substituting ##w=u'## and solving for ##u##. The resulting equation can be integrated to find the solution for ##u##, which can then be substituted back in to find the solution for ##y##.
  • #1
Turion
145
2
$$w'+2w=0\\ \frac { dw }{ dx } =-2w\\ I(x)={ e }^{ 2x }\\ \frac { dw }{ dx } { e }^{ 2x }=-2w{ e }^{ 2x }\\ \int { \frac { dw }{ dx } { e }^{ 2x } } dx=\int { -2w{ e }^{ 2x } } dx$$

Not sure what to do next.
 
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  • #2
Turion said:
$$w'+2w=0\\ \frac { dw }{ dx } =-2w\\ I(x)={ e }^{ 2x }\\ \frac { dw }{ dx } { e }^{ 2x }=-2w{ e }^{ 2x }\\ \int { \frac { dw }{ dx } { e }^{ 2x } } dx=\int { -2w{ e }^{ 2x } } dx$$

Not sure what to do next.

Starting from dw/dx = -2w, you can use the method of separation, getting w and dw on one side, and x and dx on the other. You don't need an integration factor, since the equation is already separable.

BTW, the homework template is there for a reason. Don't just delete its parts. Fair warning...
 
  • #3
You should get :

##w' + 2w = 0##
##w'e^{2x} + 2e^{2x}w = 0##

Notice the left hand side of the equation can be written :

##\frac{d}{dx} [e^{2x}w] = 0##

Try integrating now :)
 
  • #4
This first-order DE is actually the result of a reduction of a second-order DE:

$$y''-y=0\quad \quad \quad \quad \quad { y }_{ 1 }={ e }^{ x }\\ Let\quad { y }_{ 2 }=u(x){ e }^{ x }\\ Sub\quad { y }_{ 2 },\quad { y }_{ 2 }',\quad and\quad { y }_{ 2 }''\quad into\quad the\quad DE\\ { e }^{ x }{ u }''+2{ e }^{ x }u'=0\\ Let\quad w=u'\\ w'+2w=0\\ dw=-2wdx\\ \frac { -1 }{ 2w } dw=dx\\ \int { \frac { -1 }{ 2w } dw } =\int { dx } \\ \frac { -1 }{ 2 } \int { \frac { -1 }{ w } dw } =\int { dx } \\ Ignoring\quad c\quad for\quad now\\ \frac { -1 }{ 2 } ln|w|=x\\ |w|={ e }^{ -2x }\\ |\frac { du }{ dx } |={ e }^{ -2x }$$

I was wondering what I should do to proceed?
 
  • #5
Turion said:
This first-order DE is actually the result of a reduction of a second-order DE:

$$y''-y=0\quad \quad \quad \quad \quad { y }_{ 1 }={ e }^{ x }\\ Let\quad { y }_{ 2 }=u(x){ e }^{ x }\\ Sub\quad { y }_{ 2 },\quad { y }_{ 2 }',\quad and\quad { y }_{ 2 }''\quad into\quad the\quad DE\\ { e }^{ x }{ u }''+2{ e }^{ x }u'=0\\ Let\quad w=u'\\ w'+2w=0\\ dw=-2wdx\\ \frac { -1 }{ 2w } dw=dx\\ \int { \frac { -1 }{ 2w } dw } =\int { dx } \\ ***\frac { -1 }{ 2 } \int { \frac { -1 }{ w } dw } =\int { dx }*** \\ Ignoring\quad c\quad for\quad now\\ \frac { -1 }{ 2 } ln|w|=x\\ |w|={ e }^{ -2x }\\ |\frac { du }{ dx } |={ e }^{ -2x }$$

I was wondering what I should do to proceed?

I see you took the separating variables route. You made a mistake where I highlighted with ***, there are two negatives when there is only one : ##\frac { -1 }{ 2 } \int { \frac { -1 }{ w } dw } =\int { dx }##

Also, you don't need the absolute value on the logarithm in the line below it.
 

What is a first order differential equation?

A first order differential equation (ODE) is an equation that involves a function and its first derivative. It is a mathematical expression that relates the rate of change of a variable to the value of the variable itself.

What is an integrating factor?

An integrating factor is a function used to transform a first order differential equation into a form that can be easily solved by integration. It is usually a function of the independent variable in the differential equation.

How do you solve a first order differential equation using an integrating factor?

To solve a first order differential equation using an integrating factor, you first need to identify the integrating factor by multiplying the entire equation by a function of the independent variable. Then, you can integrate both sides of the equation to solve for the unknown function.

What are the benefits of using an integrating factor to solve a first order differential equation?

Using an integrating factor can make solving a first order differential equation easier and more efficient. It can also help to identify patterns and relationships between the independent and dependent variables in the equation.

What are some real-world applications of solving first order differential equations using an integrating factor?

First order differential equations and integrating factors have many applications in the fields of physics, engineering, economics, and biology. Some examples include studying population growth, modeling chemical reactions, and analyzing electrical circuits.

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